Convergence in probability does not imply convergence of the sample mean to the same value. To see this, consider a sequence of independent random variables $X_1, X_2, X_3, ...$ that each have distribution $X_k \sim k^p \cdot \text{Bern}(1/k^p)$ for some $0<p<1$. For this sequence we have:
$$\begin{matrix}
\mathbb{E}(X_k) = 1 & & & \mathbb{V}(X_k) = k^p - 1.
\end{matrix}$$
Theorem: For this sequence we have $\text{plim}_{n \rightarrow \infty} X_n = 0$ and $\text{plim}_{n \rightarrow \infty} \bar{X}_n = 1$.
Proof: We will first show that show that $\text{plim}_{n \rightarrow \infty} X_n = 0$. For any $0<\epsilon <1$ we have:
$$\lim \limits_{n \rightarrow \infty} \mathbb{P}(|X_n| > \epsilon) = \lim \limits_{n \rightarrow \infty} \mathbb{P}(X_n \geqslant 1) = \lim \limits_{n \rightarrow \infty} \frac{1}{n^p} = 0.$$
We now show that $\text{plim}_{n \rightarrow \infty} \bar{X}_n = 1$. Using Chebyshev's inequality we have:
$$\begin{equation} \begin{aligned}
\lim \limits_{n \rightarrow \infty} \mathbb{P}(|\bar{X}_n - 1| > \epsilon) &\leqslant \lim \limits_{n \rightarrow \infty} \frac{1}{\epsilon^2} \mathbb{V}(\bar{X}_n) \\
&= \lim \limits_{n \rightarrow \infty} \frac{1}{\epsilon^2} \frac{1}{n^2} \sum_{k=1}^n (k^p-1) \\
&< \lim \limits_{n \rightarrow \infty} \frac{1}{\epsilon^2} \frac{1}{n^2} \sum_{k=1}^n n^p \\
&= \lim \limits_{n \rightarrow \infty} \frac{1}{\epsilon^2} \frac{1}{n^{1-p}} =0.
\end{aligned} \end{equation}$$
This particular example uses a pathological sequence with $\lim _{n \rightarrow \infty} \mathbb{V}(X_k) = \infty$. For this sequence, there is a constant expected value, but there is convergence in probability to a different value. Perhaps the best way to get the desired result would be to impose some additional requirement like finite variance to avoid this problem.